What is the extraneous solution to these equations? $\dfrac{x^2 - 6}{x - 7} = \dfrac{8x - 13}{x - 7}$
Solution: Multiply both sides by $x - 7$ $ \dfrac{x^2 - 6}{x - 7} (x - 7) = \dfrac{8x - 13}{x - 7} (x - 7)$ $ x^2 - 6 = 8x - 13$ Subtract $8x - 13$ from both sides: $ x^2 - 6 - (8x - 13) = 8x - 13 - (8x - 13)$ $ x^2 - 6 - 8x + 13 = 0$ $ x^2 + 7 - 8x = 0$ Factor the expression: $ (x - 7)(x - 1) = 0$ Therefore $x = 7$ or $x = 1$ At $x = 7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 7$, it is an extraneous solution.